Posts: 7Member ■□□□□□□□□□
Hi, I have been working through the bryant advantage subnetting guide and understand how to manually subnet, with some exceptions. One being valid IP Address ranges. I can convert the decimal address below into binary, and its mask. I can AnD the two to get the subnet address, and convert the binary back into decimal. But, I have trouble working out the host range. Could someone explain? Using the example below?

What is the valid IP Address range for the subnet 144.45.24.0/21

144.45.24.0/21

10010000. 00101101.00011[000.00000000]

255.255.248.0/21

11111111.11111111.11111[000.00000000]

I have put brackets around the host bits. This is the bit I am stuck on. Now matter how I add them up, I do not get the answer: outlined below. Could someone explain.

All Zeros: 144.45.24.0/21
All Ones: 144.45.31.255/21

Where does the 31 come from? With respect to the brackets above. Hope someone can help.

• Posts: 95Member ■■□□□□□□□□
Phandy wrote: »
Hi, I have been working through the bryant advantage subnetting guide and understand how to manually subnet, with some exceptions. One being valid IP Address ranges. I can convert the decimal address below into binary, and its mask. I can AnD the two to get the subnet address, and convert the binary back into decimal. But, I have trouble working out the host range. Could someone explain? Using the example below?

What is the valid IP Address range for the subnet 144.45.24.0/21

144.45.24.0/21

10010000. 00101101.00011[000.00000000]

255.255.248.0/21

11111111.11111111.11111[000.00000000]

I have put brackets around the host bits. This is the bit I am stuck on. Now matter how I add them up, I do not get the answer: outlined below. Could someone explain.

All Zeros: 144.45.24.0/21
All Ones: 144.45.31.255/21

Where does the 31 come from? With respect to the brackets above. Hope someone can help.

You were very close. The valid range of address in binary are the values were the host bits are not all zero and not all ones. So...

Network:144.45.24.0
First:144.45.24.1 bin:10010000. 00101101.00011[000.00000001]
Last:144.45.31.254 bin:10010000. 00101101.00011[111.11111110]

Next:144.45.32.0

Since the first 21 bits of the address are reserved for the network that cannot change, in the third octet the first 5 bit are always 00011, when you flip the last three bits to one, you get 00011111 = 31 in decimal.
In Progress:
WGU B.S. - I.T. - Security (and all the certs that come with it)
• Posts: 7Member ■□□□□□□□□□
Thanks CIIE wannabe. I will digest all that in the morning.
• Posts: 95Member ■■□□□□□□□□
Phandy wrote: »
Thanks CIIE wannabe. I will digest all that in the morning.

You're welcome!
In Progress:
WGU B.S. - I.T. - Security (and all the certs that come with it)